# trig identities

• Jan 31st 2011, 10:50 AM
frankinaround
trig identities
Verify the given identities : (for this post the letter T will = angle theyta)

a) Sin 3T = 3 sinT - 4 sin^3T (hint 3T=2T+T)
b) cos 3T = 4cos^3T-3cosT

I used the trig identites I know to break them down, but I couldnt get it to the form of a and b. for example, for A I got this :

sin3T = sin(2T+T) = sin2t x cosT + cos2T x sinT =
1. (2sinT x cosT)cosT + (cos^2(T) - sin^2(t))sinT =
2. 2sinTcosT^2 + sin(T)cos^2(T)-sin^3(T) =
3. sin(T) (sin(T)cos^2(T) +cos^2(T)-sin^2(T))
But I get stuck at 3. Dont see how he gets 3sinT-4sin^3T as the answer. same problem with problem B. Can someone show me the light?
• Jan 31st 2011, 11:03 AM
Quote:

Originally Posted by frankinaround
Verify the given identities : (for this post the letter T will = angle theyta)

a) Sin 3T = 3 sinT - 4 sin^3T (hint 3T=2T+T)
b) cos 3T = 4cos^3T-3cosT

I used the trig identites I know to break them down, but I couldnt get it to the form of a and b. for example, for A I got this :

sin3T = sin(2T+T) = sin2t x cosT + cos2T x sinT =
1. (2sinT x cosT)cosT + (cos^2(T) - sin^2(t))sinT =
2. 2sinTcosT^2 + sin(T)cos^2(T)-sin^3(T) =
3. sin(T) (sin(T)cos^2(T) +cos^2(T)-sin^2(T))
But I get stuck at 3. Dont see how he gets 3sinT-4sin^3T as the answer. same problem with problem B. Can someone show me the light?

At line 2, you could factor out the $sinT$

$2sinTcos^2T+sinTcos^2T-sin^3T=sinT\left(2cos^2T+cos^2T\right)-sin^3T$

$=sinT\left(3cos^2T\right)-sin^3T=3sinT\left(1-sin^2T\right)-sin^3T$
• Jan 31st 2011, 12:28 PM
frankinaround
Quote:

At line 2, you could factor out the $sinT$

$2sinTcos^2T+sinTcos^2T-sin^3T=sinT\left(2cos^2T+cos^2T\right)-sin^3T$

$=sinT\left(3cos^2T\right)-sin^3T=3sinT\left(1-sin^2T\right)-sin^3T$

Hey what trig identity did you use to change 3sinTcos^2T into 3sinT(1-sin^2T) ?
• Jan 31st 2011, 12:30 PM
$cos^2T+sin^2T=1$
$\Rightarrow\ cos^2T=1-sin^2T$